# Fourier Transform of Integral of Product of Functions

• WarnK
In summary, the conversation is discussing the Fourier transform of a convolution and the differences in definitions for the transform. The speaker recommends a different definition for the transform which will make computations easier. They also mention that the answer may differ by a minus sign due to the different definitions.

#### WarnK

Hi!

I want to find the Fourier transform of

$$\int_{-\infty}^t f(s-t)g(s) ds$$.

The FT

$$\int_{-\infty}^t h(s) ds \rightarrow H(\omega)/i\omega + \pi H(0) \delta(\omega)$$

is found in lots of textbooks. So if I let h(s) = f(s-t)g(s), I need to find the FT of h(s)

$$H(\omega) = \int_{-\infty}^{\infty} h(s) \frac{e^{-i\omega s}}{\sqrt(2\pi)}ds = \int_{-\infty}^{\infty} f(s-t)g(s) \frac{e^{-i\omega s}}{\sqrt(2\pi)} ds$$.

But there I'm stumped, what can I do? I can do FT of products like f(s)g(s), but this isn't exactly like that.

Last edited:
Did you try the Fourier transform of a convolution?

Defennder said:
Did you try the Fourier transform of a convolution?

yeah, but the OP needs to reverse the "time" argument in f(.).

WarnK, where did you get that icky $1/\sqrt{2 \pi}$ definition for the F.T.?

i really recommend this definition:

$$X(f) = \int_{-\infty}^{+\infty} x(t) e^{-i 2 \pi f t} dt$$

with inverse

$$x(t) = \int_{-\infty}^{+\infty} X(f) e^{i 2 \pi f t} df$$

while remembering that $\omega \equiv 2 \pi f$, and when comparing to the double-sided Laplace Transform to substitute $i \omega \leftarrow s$.

it will make your life much easier.

I'm rather new to this myself. So the answer differs by a minus sign?